Answer:
A) ΔU = 3.9 × 10^(10) J
B) v = 8420.75 m/s
Explanation:
We are given;
Potential Difference; V = 1.3 × 10^(9) V
Charge; Q = 30 C
A) Formula for change in energy of transferred charge is given as;
ΔU = QV
Plugging in the relevant values gives;
ΔU = 30 × 1.3 × 10^(9)
ΔU = 3.9 × 10^(10) J
B) We are told that this energy gotten above is used to accelerate a 1100 kg car from rest.
This means that the initial potential energy will be equal to the final kinetic energy since all the potential energy will be converted to kinetic energy.
Thus;
P.E = K.E
ΔU = ½mv²
Where v is final velocity.
Plugging in the relevant values;
3.9 × 10^(10) = ½ × 1100 × v²
v² = [7.8 × 10^(8)]/11
v² = 70909090.9090909
v = √70909090.9090909
v = 8420.75 m/s
Answer:
C
Explanation:
The order is this: Displacement --> Velocity --> Acceleration. The velocity directly affects the displacement and the acceleration directly affects the velocity. (changing the acceleration changes the velocity which also changes the displacement).
To go forward in our order, find the slope of the curve/line. To go backwards, find the area under the curve/line. In this case, we are trying to find the displacement of the object, so we should find the area of a velocity-time graph.
Answer:
76969.29 W
Explanation:
Applying,
P = F×v............. Equation 1
Where P = Power, F = force, v = velocity
But,
F = ma.......... Equation 2
Where m = mass, a = acceleration
Also,
a = (v-u)/t......... Equation 3
Given: u = 0 m/s ( from rest), v = 12.87 m/s, t = 3.47 s
Substitute these values into equation 3
a = (12.87-0)/3.47
a = 3.71 m/s²
Also Given: m = 1612 kg
Substitute into equation 2
F = 1612(3.71)
F = 5980.52 N.
Finally,
Substitute into equation 1
P = 5980.52×12.87
P = 76969.29 W
Answer:yes
Explanation:yes it can.energy can be converted from one form to another,but cannot be created nor destroyed according to the law of conservation of energy